our good (?!) friend RootSum

*To*: mathgroup at smc.vnet.net*Subject*: [mg75022] our good (?!) friend RootSum*From*: "dimitris" <dimmechan at yahoo.com>*Date*: Sat, 14 Apr 2007 01:11:39 -0400 (EDT)

I think with the last threads it is apparent that when RootSum is generated by Integrate there are some possibilities that Mathematica may fail in the quest of a definite integration. Especially (and almost mostly) if there is involved Infinity in the integration limit(s). However this failure hasn't to do with Integrate but rather than with Limit. That is, I believe, it has to do with the difficulty of Mathematica to evaluate sometimes limits of the structure Limit[RootSum[...],x->Infinity]. For example consider the definite integral of the function [(2*Sqrt[x])/(x^3 + 3*x^2 + 2*x + 1) in the interval [0,Infinity). The integral is convergent as someone can easily proved In[791]:= ((2*Sqrt[x])/(x^3 + 3*x^2 + 2*x + 1) + O[x, #1]^3 & ) /@ {0, Infinity} Reduce[x^3 + 3*x^2 + 2*x + 1 == 0 && x >= 0, x, Reals] Out[791]= {SeriesData[x, 0, {2, 0, -4, 0, 2}, 1, 6, 2], SeriesData[x, Infinity, {2}, 5, 6, 2]} Out[792]= False Let now In[789]:= f = HoldForm[Integrate[(2*Sqrt[x])/(x^3 + 3*x^2 + 2*x + 1), x]] Mathematica can easily get an antiderivative In[796]:= g=ReleaseHold[f] D[g, x] Out[796]= 2*RootSum[1 + 2*#1^2 + 3*#1^4 + #1^6 & , (Log[Sqrt[x] - #1]*#1)/(2 + 6*#1^2 + 3*#1^4) & ] Out[797]= (2*Sqrt[x])/(1 + 2*x + 3*x^2 + x^3) However (but not surprising considering some recent posts) the definite integral is not an easy task. But it was explained this is due to the incapability of Limit and not because of Integrate algorithm. In[804]:= ReleaseHold[f /. Integrate[g_, x_] :> Integrate[g, {x, 0, Infinity}]] Out[804]= Integrate[(2*Sqrt[x])/(1 + 2*x + 3*x^2 + x^3), {x, 0, Infinity}] For this integral Mathematica finds an antiderivative and then (after verifying convergence) applies the Newton-Leibniz formula. It is obvious that the form of Newton Leibniz formula that Mathematica uses for this integral fails at Infinity as the following code shows In[805]:= Unprotect[Limit]; Limit[a___] := Null /; (Print[InputForm[limit[a]]]; False) In[807]:= ReleaseHold[f /. Integrate[g_, x_] :> Integrate[g, {x, 0, Infinity}]] (*output is ommited due to its length*) Of course in zero there is no such a problem. In[800]:= Unprotect[Limit]; Clear[Limit]; In[822]:= (Limit[g, x -> #1] & ) /@ {0, Infinity} Out[822]= {2*RootSum[1 + 2*#1^2 + 3*#1^4 + #1^6 & , (Log[-#1]*#1)/(2 + 6*#1^2 + 3*#1^4) & ], Limit[2*RootSum[1 + 2*#1^2 + 3*#1^4 + #1^6 & , (Log[Sqrt[x] - #1]*#1)/(2 + 6*#1^2 + 3*#1^4) & ], x -> Infinity]} Let see the antiderivative In[823]:= Plot[g, {x, 0, 20}] It is obvious that g->0 as x->Infinity. The question is of course WHY Limit can't recognise this limit behavior! Note that as it was pointed out in another post In[825]:= g + O[x, Infinity]^4 Series::nmer :... Series::nmer :... Out[825]= 2*RootSum[1 + 2*#1^2 + 3*#1^4 + #1^6 & , (Log[Sqrt[x] - #1]*#1)/(2 + 6*#1^2 + 3*#1^4) & ] + SeriesData[x, Infinity, {}, 4, 4, 1] Taking Limit[g->Infinity]=0 by ourselves and applying the Newton Leibniz formula results in the correct result. In[838]:= 0 - g /. x -> 0 {N[%], ReleaseHold[f /. Integrate[h_, x_] :> NIntegrate[h, {x, 0, Infinity}]]} Out[838]= -2*RootSum[1 + 2*#1^2 + 3*#1^4 + #1^6 & , (Log[-#1]*#1)/(2 + 6*#1^2 + 3*#1^4) & ] Out[839]= {0.8689140125914716+0.*I, 0.8689140125951508} As a comparison, in another CAS (due to the policy of the forum I cannot mention its name) I simply took int((2*sqrt(x))/(x^3 + 3*x^2 + 2*x + 1),x=0..infinity); / ----- \ | \ 50784 5 2760 3 22 | -4 | ) _R ln(------ _R + ---- _R - -- _R)| | / 19 19 19 | | ----- | \_R = %1 / 6 4 2 %1 := RootOf(33856 _Z + 1104 _Z + 32 _Z + 1) evalf(%); 0.8689140124 - 0. I Very good performance indeed. In the same system I got int((2*sqrt(x))/(x^3 + 3*x^2 + 2*x + 1),x); / ----- 5 3 \ | \ 1/2 50784 _R 2760 _R 22 _R | 4 | ) _R ln(x - --------- + -------- - -----)| | / 19 19 19 | | ----- | \_R = %1 / 6 4 2 %1 := RootOf(33856 _Z + 1104 _Z + 32 _Z + 1) simplify(diff(%,x)); 1/2 2 x ------------------- 3 2 x + 3 x + 2 x + 1 In the field of integration I do consider Mathematica by far more superior than other CASs but I would like if someone could explain me why, even though both Mathematica and the other CAS I use, generating a quite similar root sum object Mathematica has such a low performance. Note that in all of my recent posts dealing with integrals of rational functions in which Integrate generates a RootSum and Limit fails to get the limit behavior of the antiderivative at Infinity (ussually zero) the other CAS simply do not face problems.